For the consecutive first-order reactions: $A\overset{\quad K_{1}\quad}{\rightarrow}B\overset{\quad

Question

For the consecutive first-order reactions: $A\overset{\quad K_{1}\quad}{\rightarrow}B\overset{\quad K_{2}\quad}{\rightarrow}C$, in what condition, $\lbrack C\rbrack = \left\lbrack A_{0} \right\rbrack.\left( 1 - e^{- K_{1}t} \right)$?

Accepted Answer

On derivation $\lbrack Cbrack = \left\lbrack A_{0} ightbrack\left\lbrack 1 - \frac{K_{2}.e^{- K_{1}t} - K_{1}.e^{- K_{2}t}}{K_{2} - K_{1}} ightbrack\ \left( 	ext{when }K_{1} 
eq K_{2} ight)$If K1 << K2, [C] becomes $\left\lbrack A_{0} ightbrack.\left( 1 - e^{- K_{1}t} ight)$

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